What is the solution for 2y^{''}+3y^'=0,y(0)=1,y^'(0)=1 ?

The solution for 2y^{''}+3y^'=0,y(0)=1,y^'(0)=1 is y=1+2/3-2/3 e^{-(3t)/2}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2y^{''}+3y^'=0,y(0)=1,y^'(0)=1

2 y^{^{''}} + 3 y^{^{'}} = 0, y (0) = 1, y^{^{'}} (0) = 1

y = 1 + \frac{2}{3} - \frac{2}{3} e^{- \frac{3 t}{2}}

Solution steps

2 y^{''} (t) + 3 y^{'} (t) = 0

Solve linear ODE:

y = 1 + \frac{2}{3} - \frac{2}{3} e^{- \frac{3 t}{2}}

y = 1 + \frac{2}{3} - \frac{2}{3} e^{- \frac{3 t}{2}}

x^{^{''}} + 8 x^{^{'}} = 0, x (0) = - 3, x^{^{'}} (0) = 5

y^{^{''}} - 6 y^{^{''}} + 8 y = 0

y^{^{''}} - 14 y^{^{'}} + 113 y = 0

y^{^{''}} - 6 y^{^{'}} + 15 y = 0

55 y^{^{''''}} + 136 y^{^{''}} + 25 y = 0

What is the solution for 2y^{''}+3y^'=0,y(0)=1,y^'(0)=1 ?
The solution for 2y^{''}+3y^'=0,y(0)=1,y^'(0)=1 is y=1+2/3-2/3 e^{-(3t)/2}